Question

A ship leaves port with a bearing of N 55 degrees W. After traveling 16 miles, the ship then turns 90 degrees and travels on a bearing of S 35 degrees W for 5 miles. At that time, what is the bearing of the ship from port? Round to the nearest tenth.

Answer 1

Answer on Question #81506 – Math – Trigonometry

Question

A ship leaves port with a bearing of N 55 degrees W. After traveling 16 miles, the ship then turns 90 degrees and travels on a bearing of S 35 degrees W for 5 miles. At that time, what is the bearing of the ship from port? Round to the nearest tenth.

Solution

Fig. 1

We have right triangle as showing at Fig.1

$R_{1}$ - first displacement, equal 16 mi with a bearing of N 55 degrees W.

$R_{2}$ - second displacement, equal 5 mi with a bearing of S 35 degrees W, after the ship then turns 90 degrees.

Then we obtain a right triangle. $R_{1}$ and $R_{2}$ are catheti, in right triangle (ABC). $\angle B = 90{}^{\circ}$

Mark this angle $x$, then:

$$x = 55{}^{\circ} + \angle BAC$$ $$\angle BAC = \tan^{-1} \left(\frac{R_{2}}{R_{1}}\right) = \tan^{-1} \left(\frac{5}{16}\right) = \tan^{-1} (0.3125) \approx 17.4{}^{\circ}$$ $$x = 55{}^{\circ} + \angle BAC = 72.35{}^{\circ}$$

**Answer:** The bearing of the ship from port is equal N $72.4{}^{\circ}$ W.

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